ELECTRIC POTENTIAL
Electric potential defind as the amount of work done bringing a test charge from infinite
to a particular point inside the electric field .
1. Mathematicaly, electric potential (v)=w/q0
2. It is a scalar quantity.
3. SI unit =j/c or Nm/C or volt
4. E.S.U unit =stat -volt
5. E.M.U unit = ab-volt
6. Dimension = [V] = [W] /[ q]
= [M¹L²T -3 ] /[A¹T¹]
= [M¹L²T -3 A -1]
RELATION BETWEEN VOLT & STAT VOLT…….
1 volt =1j/1c
= 10⁷erg/3×10⁹ stat coulumb
=1/3×10² stat volt
● 1 volt = 1/ 300 stat volt
RELATION BETWEEN VOLT & ABVOLT……
1 volt= 1j/1c
= 10⁷erg /( 1/10 ab volt)
● 1 volt = 10⁸ ab-volt
1
, ELECTRIC POTENTIAL AT A POINT DUE TO A POINT CHARGE ….
Consider a point charge q creates are electric field E vector
Let a test charge q0is brought from infinite to a particular point p inside the electric
field .
Let the test charge move a small displacement dx due to force F .
So a small work dw is done
dw = F vector. dx vector
=F dx cos ∅
= F dx 180⁰
= - F ds —-------------------(1)
The electric field force between q and q0 is given by F = 1/4πE0 q q0/x²
Using this value in equation1 we get dw= -1/4πE0 q q0/x² dx
Total work done ,integration of dw
(Upper limits w and lower limit 0)
Integration of 1/4πE0 . q q0/x² dx
(Upper limit r and lower limit ∞)
q q0/4πE0 indignation of x -2 dx
(Upper limit r & lower limit ∞)
-qq0/4πE0 [ x -2+1/-2+1]
2
Electric potential defind as the amount of work done bringing a test charge from infinite
to a particular point inside the electric field .
1. Mathematicaly, electric potential (v)=w/q0
2. It is a scalar quantity.
3. SI unit =j/c or Nm/C or volt
4. E.S.U unit =stat -volt
5. E.M.U unit = ab-volt
6. Dimension = [V] = [W] /[ q]
= [M¹L²T -3 ] /[A¹T¹]
= [M¹L²T -3 A -1]
RELATION BETWEEN VOLT & STAT VOLT…….
1 volt =1j/1c
= 10⁷erg/3×10⁹ stat coulumb
=1/3×10² stat volt
● 1 volt = 1/ 300 stat volt
RELATION BETWEEN VOLT & ABVOLT……
1 volt= 1j/1c
= 10⁷erg /( 1/10 ab volt)
● 1 volt = 10⁸ ab-volt
1
, ELECTRIC POTENTIAL AT A POINT DUE TO A POINT CHARGE ….
Consider a point charge q creates are electric field E vector
Let a test charge q0is brought from infinite to a particular point p inside the electric
field .
Let the test charge move a small displacement dx due to force F .
So a small work dw is done
dw = F vector. dx vector
=F dx cos ∅
= F dx 180⁰
= - F ds —-------------------(1)
The electric field force between q and q0 is given by F = 1/4πE0 q q0/x²
Using this value in equation1 we get dw= -1/4πE0 q q0/x² dx
Total work done ,integration of dw
(Upper limits w and lower limit 0)
Integration of 1/4πE0 . q q0/x² dx
(Upper limit r and lower limit ∞)
q q0/4πE0 indignation of x -2 dx
(Upper limit r & lower limit ∞)
-qq0/4πE0 [ x -2+1/-2+1]
2
